Leetcode: Sum Root to Leaf Numbers

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

For example,

    1
   / \
  2   3

The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.

Return the sum = 12 + 13 = 25.

Understand the problem:
The problem asks for the sum of all root-to-leaf numbers. The problem itself is not hard to understand. Use DFS is the natural way. The only thing to note is the overflow problem, so we may use a long long to store the intermediate results.

Solution:

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

/**  * Definition for binary tree  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode(int x) { val = x; }  * }  */ public class Solution {     long pathSum = 0;     public int sumNumbers(TreeNode root) {           sumNumbersHelper(root, 0);                   return (int)pathSum;     }           private void sumNumbersHelper(TreeNode root, long curSum) {         if (root == null) {             return;         }                   curSum = curSum * 10 + root.val;                   if (root.left == null && root.right == null) {             pathSum += curSum;         }           sumNumbersHelper(root.left, curSum);         sumNumbersHelper(root.right, curSum);     } }

Update on 10/8/14:

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

/**  * Definition for binary tree  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode(int x) { val = x; }  * }  */ public class Solution {     public int sumNumbers(TreeNode root) {           return sumNumbersHelper(root, 0);     }           private int sumNumbersHelper(TreeNode root, int preSum) {         if (root == null) {             return 0;         }                   int curSum = root.val + preSum * 10;                   if (root.left == null && root.right == null) {             return curSum;         }                   return sumNumbersHelper(root.left, curSum) + sumNumbersHelper(root.right, curSum);     } }